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Question Number 121085 by TITA last updated on 05/Nov/20

Commented by TITA last updated on 05/Nov/20

please help

$${please}\:{help} \\ $$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

3x^2 +x+2=0⇒x=((−1±i(√(23)))/6) α+β=−(1/3) αβ=(2/3) ((α+β)/(αβ))=((−1)/2) ⇒(1/α)+(1/β)=−(1/2)⇒ (1/α^2 )+(1/β^2 )+(2/(αβ))=(1/4) (1/α^2 )+(1/β^2 )=(1/4)−3=−((11)/4) Equation x^2 −((1/α^2 )+(1/β^2 ))x+((1/(αβ)))^2 =x^2 +((11)/4)x+(9/4)=0 4x^2 +11x+9=0

$$\mathrm{3}{x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0}\Rightarrow{x}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{23}}}{\mathrm{6}}\: \\ $$$$\alpha+\beta=−\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\alpha\beta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{\alpha+\beta}{\alpha\beta}=\frac{−\mathrm{1}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }+\frac{\mathrm{2}}{\alpha\beta}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{3}=−\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$ \\ $$$${Equation}\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right){x}+\left(\frac{\mathrm{1}}{\alpha\beta}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}}{x}+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{9}=\mathrm{0} \\ $$

Answered by liberty last updated on 05/Nov/20

(14) 3x^2 +x+2=0 → { (α),(β) :} → { ((α=((−1+(√(1−24)))/6) = ((−1+i(√(23)))/6))),((β=((−1−i(√(23)))/6))) :} → { ((α^2 =((−22−2i(√(23)))/(36))= ((−11−i(√(23)))/(18)))),((β^2 =((−11+i(√(23)))/(18)))) :} → { (((1/α^2 ) = ((−18)/(11+i(√(23)))) = ((−11+i(√(23)))/8))),(((1/β^2 )=((−18)/(11−i(√(23))))=((−11−i(√(23)) )/8))) :}

$$\left(\mathrm{14}\right)\:\mathrm{3x}^{\mathrm{2}} +\mathrm{x}+\mathrm{2}=\mathrm{0}\:\rightarrow\begin{cases}{\alpha}\\{\beta}\end{cases} \\ $$$$\:\rightarrow\:\begin{cases}{\alpha=\frac{−\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{24}}}{\mathrm{6}}\:=\:\frac{−\mathrm{1}+{i}\sqrt{\mathrm{23}}}{\mathrm{6}}}\\{\beta=\frac{−\mathrm{1}−{i}\sqrt{\mathrm{23}}}{\mathrm{6}}}\end{cases} \\ $$$$\rightarrow\begin{cases}{\alpha^{\mathrm{2}} =\frac{−\mathrm{22}−\mathrm{2}{i}\sqrt{\mathrm{23}}}{\mathrm{36}}=\:\frac{−\mathrm{11}−{i}\sqrt{\mathrm{23}}}{\mathrm{18}}}\\{\beta^{\mathrm{2}} =\frac{−\mathrm{11}+{i}\sqrt{\mathrm{23}}}{\mathrm{18}}}\end{cases} \\ $$$$\rightarrow\begin{cases}{\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:=\:\frac{−\mathrm{18}}{\mathrm{11}+{i}\sqrt{\mathrm{23}}}\:=\:\frac{−\mathrm{11}+{i}\sqrt{\mathrm{23}}}{\mathrm{8}}}\\{\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{−\mathrm{18}}{\mathrm{11}−{i}\sqrt{\mathrm{23}}}=\frac{−\mathrm{11}−{i}\sqrt{\mathrm{23}}\:}{\mathrm{8}}}\end{cases} \\ $$

Answered by TANMAY PANACEA last updated on 05/Nov/20

α+β=((−1)/3) αβ=(2/3) 3x^2 +x+2=0 x=((−1±(√(1−4×3×2)))/(2×3))=((−1±i(√(23)))/6) α=((−1+i(√(23)))/6) β=((−1−i(√(23)))/6) (1/α^2 )=((1−i×2(√(23)) −23)/(36))=((−22−i(√(23)))/(36)) (1/β^2 )=((1+i2(√(23)) −23)/(36))=((−22+i2(√(23)))/(36)) eqn x^2 −x((1/α^2 )+(1/β^2 ))+(1/((αβ)^2 ))=0 x^2 −x{(((α+β)^2 −2αβ)/((αβ)^2 ))}+(1/((αβ)^2 ))=0 x^2 −x((((1/9)−2×(2/3))/(4/9)))+(9/4)=0 x^2 −x(((1−12)/4))+(9/4)=0 4x^2 +11x+9=0 27α^4 =11α+10 α=((−1+i(√(23)))/6)→α^2 =((1−i2(√(23)) −23)/(36))=((−22−i2(√(23)))/(36))=((−11−i(√(23)))/(18)) α^4 =((121+i22(√(23)) −23)/(324))=((98+i22(√(23)))/(324))=((49+i11(√(23)))/(162)) 27α^4 =27×((49+i11(√(23)))/(162))=((49+i11(√(23)))/6) 11α+10 11×((−1+i(√(23)))/6)+10 ((−11+i11(√(23)) +60)/6) =((49+i11(√(23)))/6) so LHS=RHS

$$\alpha+\beta=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\alpha\beta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}×\mathrm{3}×\mathrm{2}}}{\mathrm{2}×\mathrm{3}}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{23}}}{\mathrm{6}} \\ $$$$\alpha=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{23}}}{\mathrm{6}}\:\:\:\beta=\frac{−\mathrm{1}−{i}\sqrt{\mathrm{23}}}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }=\frac{\mathrm{1}−{i}×\mathrm{2}\sqrt{\mathrm{23}}\:−\mathrm{23}}{\mathrm{36}}=\frac{−\mathrm{22}−{i}\sqrt{\mathrm{23}}}{\mathrm{36}} \\ $$$$\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{\mathrm{1}+{i}\mathrm{2}\sqrt{\mathrm{23}}\:−\mathrm{23}}{\mathrm{36}}=\frac{−\mathrm{22}+{i}\mathrm{2}\sqrt{\mathrm{23}}}{\mathrm{36}} \\ $$$${eqn} \\ $$$${x}^{\mathrm{2}} −{x}\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\left(\alpha\beta\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}\left\{\frac{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta}{\left(\alpha\beta\right)^{\mathrm{2}} }\right\}+\frac{\mathrm{1}}{\left(\alpha\beta\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}\left(\frac{\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{2}×\frac{\mathrm{2}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{9}}}\right)+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}\left(\frac{\mathrm{1}−\mathrm{12}}{\mathrm{4}}\right)+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{27}\alpha^{\mathrm{4}} =\mathrm{11}\alpha+\mathrm{10} \\ $$$$\alpha=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{23}}}{\mathrm{6}}\rightarrow\alpha^{\mathrm{2}} =\frac{\mathrm{1}−{i}\mathrm{2}\sqrt{\mathrm{23}}\:−\mathrm{23}}{\mathrm{36}}=\frac{−\mathrm{22}−{i}\mathrm{2}\sqrt{\mathrm{23}}}{\mathrm{36}}=\frac{−\mathrm{11}−{i}\sqrt{\mathrm{23}}}{\mathrm{18}} \\ $$$$\alpha^{\mathrm{4}} =\frac{\mathrm{121}+{i}\mathrm{22}\sqrt{\mathrm{23}}\:−\mathrm{23}}{\mathrm{324}}=\frac{\mathrm{98}+{i}\mathrm{22}\sqrt{\mathrm{23}}}{\mathrm{324}}=\frac{\mathrm{49}+{i}\mathrm{11}\sqrt{\mathrm{23}}}{\mathrm{162}} \\ $$$$\mathrm{27}\alpha^{\mathrm{4}} =\mathrm{27}×\frac{\mathrm{49}+{i}\mathrm{11}\sqrt{\mathrm{23}}}{\mathrm{162}}=\frac{\mathrm{49}+{i}\mathrm{11}\sqrt{\mathrm{23}}}{\mathrm{6}} \\ $$$$\mathrm{11}\alpha+\mathrm{10} \\ $$$$\mathrm{11}×\frac{−\mathrm{1}+{i}\sqrt{\mathrm{23}}}{\mathrm{6}}+\mathrm{10} \\ $$$$\frac{−\mathrm{11}+{i}\mathrm{11}\sqrt{\mathrm{23}}\:+\mathrm{60}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{49}+{i}\mathrm{11}\sqrt{\mathrm{23}}}{\mathrm{6}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{L}}{HS}={RHS} \\ $$$$ \\ $$

Commented by TITA last updated on 05/Nov/20

thanks sir

$${thanks}\:{sir} \\ $$

Commented by TANMAY PANACEA last updated on 05/Nov/20

most welcome

$${most}\:{welcome} \\ $$

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